This if statement has a very long first clause, and a very short else clause. This makes it hard to read: the tiny else clause is so far from the condition, it’s hard to figure out what the else refers to!
if (iptStoad() && !(miua() || ma == 5) && iqod()) {
...
...
// Pretend there is lots of code here
...
...
} else {
isass();
}
Improve readability by refactoring this conditional so that its two clauses are swapped: what is now the second clause (the else clause) comes first, and the first clause comes second.
if (!iqod() || miua() || ma == 5 || !iptStoad()) {
isass();
} else {
...
...
// Pretend there is lots of code here
...
...
}
Things to double-check in your solution:
!(...) Instead, make sure you negate the condition by changing each part of it.Pretend there is lots of code here when you write out your solution! Just draw three dots; that’s enough.Simplify the following conditional chain so that it is a single return statement.
if (orme == 3 && !ri || eshGiome() || drau() && !ri || eshGiome()) {
if (eshGiome()) {
if (!ri) {
return true;
}
}
if (!psem) {
return true;
}
}
return false;
return (!psem || orme == 3 || drau()) && (!ri || eshGiome());
Bonus challenge: rewrite the if/else chain above so that instead of consisting of many return true; statements with one return false; at the end, it has many return false; statements with one return true; at the end.
if (!drau() && orme != 3 && psem) {
if (ri) {
return false;
}
if (!eshGiome()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (e < 4) {
rinel();
}
if (ipto == true && e > 4) {
ralUman();
}
if (a > 6 && e > 4 && ipto != true) {
prar();
} else if (nau == false && e > 4 && ipto != true && a < 6) {
zoiTaqan();
}
{
if (e < 4) {
rinel();
}
if (ipto) {
ralUman();
}
if (a > 6) {
prar();
}
if (!nau) {
zoiTaqan();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: